Question

1. a student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. the cliff is 50 m above a flat, horizontal beach as shown in figure p3.7. a. what are the coordinates of the initial position of the stone? b. what are the components of the initial velocity? c. write the equations for the (v_x -)and (v_y -)components of the velocity of the stone with time. d. write the equations for the position of the stone with time, using the coordinates in figure p3.7. e. how long after being released does the stone strike the beach below the cliff? f. with what speed and angle of impact does the stone land?

Explanation:

Step1: Define coordinate system

Assume the student - throwing point as the origin $(0,0)$ with the positive $x$ - direction as the horizontal direction of the throw and the positive $y$ - direction as the upward direction.

Step2: Find initial position coordinates

The initial position of the stone is at the origin of our coordinate system. So the coordinates are $(x_0,y_0)=(0,0)$.

Step3: Determine initial - velocity components

The stone is thrown horizontally with a speed of $v_0 = 18\ m/s$. The initial vertical velocity $v_{0y}=0\ m/s$ and the initial horizontal velocity $v_{0x}=18\ m/s$.

Step4: Write velocity - time equations

The horizontal component of velocity is constant because there is no acceleration in the horizontal direction ($a_x = 0$). So $v_x=v_{0x}=18\ m/s$. The vertical component of velocity is affected by gravity ($a_y=-g=- 9.8\ m/s^2$). Using the equation $v = v_0+at$, we have $v_y=v_{0y}-gt=-gt$.

Step5: Write position - time equations

The horizontal position is given by $x = x_0+v_{0x}t=18t$ (since $x_0 = 0$). The vertical position is given by $y=y_0+v_{0y}t-\frac{1}{2}gt^2=-\frac{1}{2}gt^2$ (since $y_0 = 0$ and $v_{0y}=0$).

Step6: Find time of flight

When the stone strikes the beach, $y=- 50\ m$. Using $y =-\frac{1}{2}gt^2$, we can solve for $t$. Rearranging the equation gives $t=\sqrt{\frac{-2y}{g}}$. Substituting $y=-50\ m$ and $g = 9.8\ m/s^2$, we have $t=\sqrt{\frac{2\times50}{9.8}}\approx3.19\ s$.

Step7: Find speed and angle of impact

First, find the vertical component of velocity at impact. Substitute $t = 3.19\ s$ into $v_y=-gt$, so $v_y=-9.8\times3.19=-31.26\ m/s$. The horizontal component of velocity $v_x = 18\ m/s$. The speed of impact $v=\sqrt{v_x^2 + v_y^2}=\sqrt{18^2+( - 31.26)^2}\approx36.0\ m/s$. The angle of impact $\theta=\tan^{-1}(\frac{v_y}{v_x})=\tan^{-1}(\frac{-31.26}{18})\approx - 60.0^{\circ}$ (the negative sign indicates the angle is below the horizontal).

Answer:

a. $(0,0)$
b. $v_{0x}=18\ m/s$, $v_{0y}=0\ m/s$
c. $v_x = 18\ m/s$, $v_y=-gt$
d. $x = 18t$, $y=-\frac{1}{2}gt^2$
e. $t\approx3.19\ s$
f. Speed: $v\approx36.0\ m/s$, Angle: $\theta\approx - 60.0^{\circ}$